i was trying to attend the Mathematica seminars last month to see if can
build a matrix matching model using this new tool (after searching the group
archive and could not find a close one). i had to reschedule the seminars in
the next 10 days because some technical issue kept refusing me into the classroom...
Before I can get a better sense with Mathematica, can anyone tell me if a
basic matrix matching model close to described below does exist in public or
can be built handily?
Scenario 1:
Two m X n matrices A and B, where first column are items (represented by
names or IDs, e.g. computer engineering or
88888888), last column are importance level (0~10), and rest of the columns
are descriptors (interpreted as numbers,
e.g. 0%~100%, 0~ 20,000 km, 0 or 1, which can be standardized).
Matching these two matrices A and B to determine their closeness (or level
of matching), in one way, by comparing
the descriptors of each item and calculating their overall distance,
adjusted by the levels of importance:
D1 = SUM{SQUARE[L1(B) - L1(A)]+ SQUARE[L2(B) -L2(A)] +...+
SQUARE[Ln(B)-Ln(A)]}*IM1^2
D2 = SUM{SQUARE[L1(B) - L1(A)]+ SQUARE[L2(B) -L2(A)] +...+
SQUARE[Ln(B)-Ln(A)]}*IM2^2
Dm = SUM{SQUARE[L1(B) - L1(A)]+ SQUARE[L2(B) -L2(A)] +...+
SQUARE[Ln(B)-Ln(A)]}*IMm^2
D = (D1 + D2 +... + Dm)/m
D: overall distance
L: quantified and standardized level of descriptor
IM: level of importance
Alternative Scenario:
Two m X n matrices A and B, where first column are items (represented by
names or IDs, e.g. computer engineering or 88888888) and rest of the columns
are descriptors (interpreted as numbers, e.g. 0~10, 0%~100%, 0~ 20,000 km, 0
or 1, which can be standardized).
Matching these two matrices A and B to determine their closeness (or level
of matching) by comparing the descriptors of each item and calculating their
overall distances:
D1 = SUM{SQUARE[L1(B) - L1(A)]+ SQUARE[L2(B) -L2(A)]
+...+ SQUARE[Ln(B)-Ln(A)]}^2
D2 = SUM{SQUARE[L1(B) - L1(A)]+ SQUARE[L2(B) -L2(A)]
+...+ SQUARE[Ln(B)-Ln(A)]}^2
Dm = SUM{SQUARE[L1(B) - L1(A)]+ SQUARE[L2(B) -L2(A)]
+...+ SQUARE[Ln(B)-Ln(A)]}^2
D = (D1 + D2 +... + Dm)/m
D: overall distance
L: quantified and standardized level of descriptor
Many thanks,
Sophie